Applications of Derivatives
Related Rates
General steps
1. Draw a picture!! (This may not be possible for every problem, but there’s usually something you can
draw.)
2. Label everything. If a quantity is fixed for the entire problem, write in the number. If it can change,
then assign it a variable. There are often multiple ways to draw and label things, but the final answer
will be the same irrespective of how you label things.
3. Write down what you know, and what you want to know. Note: When writing down given/known rates
of change, make them positive if the variable is getting larger, negative if the variable is getting smaller
(this is going to depend on how you labeled your picture).
4. Figure out how everything is related and come up with a formula relating the variables. This can involve
using geometric formulas, triangles, similar triangles, etc.
5. Differentiation implicitly with respect to t (or whatever the independent variable is). Remember, in
d
terms from the chain rule
these problems all variables are viewed as functions of t, so you’ll pick up dt
as you differentiate.
6. Plug in known values for variables and rates, then solve for the quantity in which you’re interested.
Warning: DO NOT plug in numbers for quantities that can change until after you differentiate!! If a
quantity is constant throughout the entire problem and cannot change, then you should have already
put in the picture as a number in step 2. (Replacing a variable with an expression involving another
variable is allowed.)
Note: It’s okay to get a negative answer in these problems. A negative answer tells us that that the quantity
is decreasing and positive tells us it’s increasing. On a test or homework I want to see the correct sign in your
answer. In some books (not ours) all rates are positive in the final answers, even if the quantity is decreasing
(the negative may be assumed from inclusion of the word “decreasing” in the problem).
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Problems
1. A spotlight on the ground shines on a building 36 ft away. A man 6 ft tall walks from the spotlight
toward the building at a speed of 10 ft/s. How fast is the length of his shadow on the building changing
when he is 24 ft from the building?
2. Each side of a square is increasing at a rate of 6 cm/s. At what rate is the area of the square increasing
when the area of the square is 16 cm2 ?
3. A point is moving along the graph of y =
for each value of x.
2
such that dx/dt is 4 centimeters per minute. Find dy/dt
3+x
4. A rectangular swimming pool is being filled with water at a rate of 5 m3 /min. The length of the pool
is 10 m and the width is 4 m. How fast is the height of the water increasing?
5. Car A is 100 mi. east of Car B at 3:00pm. Car A is moving west at 60 mi/h and Car B is moving south
at 70 mi/h. How fast is the distance between the cars changing at 4:00pm?
6. A balloon rises at a rate of 3 m/s from a point on the ground 30 m from an observer. Find the rate of
change of the distance between the observer and the balloon when the balloon is 30 m above the ground.
7. A company that manufactures sport supplements calculates that its costs and revenue can be modeled
by the equations
1
C = 125,000 + 0.75x
and
R = 250x − x2
10
where x is the number of units of sport supplements produced in 1 week. If production in one particular
week is 1000 units and is increasing at a rate of 150 units per week find:
(a) The rate at which the cost is changing.
(b) The rate at which the revenue is changing.
(c) The rate at which the profit is changing.
2
Answers
1. −2.1 f t/s
2. 48 cm2 /s
3.
dy
dt
8
= − (3+x)
2
4. + 18 m/min
5. +31 mi/h
6. The picture we drew in class formed a right triangle with base 30, height y, and hypotenuse x, so the
equation relating them is x2 = 302 + y 2 . We are given that dy
= +3 , and we want to know what dx
dt
√
√ dt
is when y = 30 (note: when y = 30, x = 302 + 302 = 1800 ≈ 42.43). Differentiating the equation
yields
dy
dx
= 0 + 2y
2x
dt
dt
Plugging in all the values we know and solving for
2 · 42.43 ·
dx
dt
gives us:
dx
= 2 · 30 · 3
dt
dx
90
=
≈ 2.12 m/s
dt
42.43
7. For all parts, we’re finding the rates for the particular week referenced in the problem, when x = 1000.
There aren’t any pictures to draw for this problem, and the equations are already given to us. We’re
also given that dx
dt = +150. So, all we need to do is differentiate C, R, and
P = R − C = 250x −
and plug in our values for x and
1 2
1
x − (125000 + 0.75x) = − x2 + 249.25x − 125000
10
10
dx
dt :
(a)
dx
1
dx
dR
1
dR
= 250 ·
−
·2·x·
−→
= 250 · 150 − · 1000 · 150 = 7500
dt
dt
10
dt
dt
5
(b)
dC
dx
dC
= 0.75 ·
−→
= 0.75 · 150 = 112.5
dt
dt
dt
(c)
1
dx
dx
dP
1
dP
=− ·2·x·
+ 249.25 ·
−→
= − · 2 · 1000 · 150 + 249.25 · 150 = 7387.5
dt
10
dt
dt
dt
10
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Increasing and Decreasing Functions
Definition 1. Let f (x) be a function defined on an interval I.
• f (x) is said to be increasing on I if f (x1 ) ≤ f (x2 ) for any two numbers x1 , x2 in I such that x1 < x2 .
Informally, this means that we’re moving uphill (as we travel in the positive x direction).
• f (x) is said to be decreasing on I if f (x1 ) ≥ f (x2 ) for any two numbers x1 , x2 in I such that x1 < x2 .
Informally, this means that we’re moving downhill (as we travel in the positive x direction).
(a) f (x) is increasing
(b) f (x) is increasing
(c) f (x) is decreasing
(d) f (x) is decreasing
Figure 1: Some Examples of Increasing and Decreasing Functions
Definition 2 (Critical Numbers). A critical number (or critical value) c of a function f (x) is any number [in
the domain of f ] such that f ′ (c) = 0 or f ′ (c) is undefined.
Theorem 3 (“Test for Increasing and Decreasing Functions”). Let f (x) be a differentiable function on the
interval I,
1. If f ′ (x) > 0 for all x in I, then f (x) is increasing on I.
2. If f ′ (x) < 0 for all x in I, then f (x) is decreasing on I.
3. If f ′ (x) = 0 for all x in I, then f (x) is constant on I.
How we use this
Finding intervals on which a function is increasing, and on which it’s decreasing.
1. Find all critical numbers (x-values where f ′ is zero or undefined). . .
2. . . . and plot them on a number line. This divides the number line into intervals.
3. Pick a test value from each interval (I’ll denote these with a star) and plug into f ′ to see if the derivative
is (+) or (−). Note: we don’t need the actual value of f ′ at each test value - only the sign.
4. Interpret:
• f is increasing on the (+) intervals.
• f is decreasing on the (−) intervals.
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Extrema
Definition 4. Let f (x) be defined at x = c.
1. f (c) is said to be a relative maximum (or local maximum) of f if and only if f (c) is greater than the
surrounding function values, i.e., f (c) is the largest function value “close to” x = c. (Looks like the
peak of a mountain.)
2. f (c) is said to be a relative minimum (or local minimum) of f if and only if f (c) is less than the
surrounding function values, i.e., f (c) is the smallest function value “close to” x = c. (Looks like the
bottom of a valley.)
Definition 5. Let f (x) be defined on an interval I containing c.
• f (c) is said to be the absolute maximum of f on I if and only if f (c) ≥ f (x) for all x in I.
• f (c) is said to be the absolute minimum of f on I if and only if f (c) ≤ f (x) for all x in I.
How we use this
Finding the relative maxima and minima of a function.
Theorem 6 (The First Derivative Test). Suppose c is a critical number of a [continuous] function f ,
1. If f ′ changes from (+) → (−) (i.e., it goes from increasing to decreasing) at c, then f has a relative
maximum at (c, f (c))
2. If f ′ changes from (−) → (+) (i.e., it goes from decreasing to increasing) at c, then f has a relative
minimum at (c, f (c))
3. If f ′ does not change sign (i.e., we have (+) → (+) or (−) → (−)) or if f (c) is undefined, then f has
no relative maximum or minimum at c.
Finding the absolute maximum and absolute minimum of a continuous function y = f (x) on a
closed interval [a, b].
0. Check to make sure the interval is closed (easy) and that the function is continuous on the interval (it’s
okay to have discontinuities outside the interval).
1. Find all critical values of f in the interval (we don’t care about those that aren’t in the interval).
2. Evaluate f (not f ′ ) at those critical points and at a and b (the endpoints of the interval).
3. The largest value from step 2 is the absolute maximum of f on [a, b], the smallest value is the absolute
minimum of f on [a, b].
Notes: When finding absolute maximum/minimum, we’re interested in the function values (y’s); for many
other problems in this chapter, we’re more interested in the x-values. The absolute maximum or minimum
may occur at more than one x-value.
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Concavity
Definition 7 (Formal definition of concavity). Let f be a differentiable function on some open interval I.
1. f is said to be concave up on I if f ′ is an increasing function on I.
2. f is said to be concave down on I if f ′ is a decreasing function on I.
Definition 8 (Informal definition of concavity). “Concave up, like a cup. Concave down, like a frown.”
Definition 9. A point where f changes from concave up to concave down or from concave down to concave
up is called an inflection points or a point of inflection (or inflection point).
(a) A concave up function
(b) A concave down function
Figure 2: Concavity Examples
(a) Inc. and Concave Up
(b) Inc. and Concave Down
(c) Dec. and Concave Up
(d) Dec. and Concave Down
Theorem 10 (“Concavity Test” - this is what some people mean by “The Second Derivative Test”).
• If f ′′ (x) > 0 for all x in an interval I, then f (x) is concave up on I.
• If f ′′ (x) < 0 for all x in an interval I, then f (x) is concave down on I.
Theorem 11 (The Second Derivative Test). Suppose f ′′ (x) is continuous near c,
1. If f ′ (c) = 0 and f ′′ (c) > 0, then f has a local minimum at c.
2. If f ′ (c) = 0 and f ′′ (c) < 0, then f has a local maximum at c.
3. If f ′′ (c) = 0, or if f ′ (c) or f ′′ (c) is undefined, then this test is inconclusive.
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How we use this
Finding local maxima and minima.
The Second Derivative Test gives us another way of checking to see if a critical value corresponds to a local
maximum or local minimum.
Finding the Point of Diminishing Returns.
If f changes from concave up to concave down at an inflection point, then (in some contexts) that inflection
point is called the “point of diminishing returns”.
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Graphical Example
4
3
2
1
-6
-4
2
-2
4
6
8
-1
-2
(Most of the following numbers are approximate and have been rounded to 2 decimal places.)
The coordinates of the relative maxima are: (−5.52, 4.97), (−2.29, 1.90), (1.58, 0.61), (5.97, 1.68).
The coordinates of the relative minima are: (−4, 0), (0, 0), (3.92, −1.55), (7, 0).
The coordinates of the inflection points are: (−5.01, 3.07), (−3.29, 0.84), (−1.27, 1.02), (0.8, 0.3), (2.9, −0.56),
(5.09, 0.24), (6.61, 0.69).
This function is increasing on the intervals (−∞, −5.52), (−4, −2.29), (0, 1.58), (3.92, 5.97), (6.61, ∞).
This function is decreasing on the intervals (−5.52, −4), (−2.29, 0), (1.58, 3.92), (5.97, 6.61).
This function is concave up on the intervals (−5.01, −3.29), (−1.27, 0.8), (2.9, 5.09), (6.61, ∞).
This function is concave down on the intervals (−∞, −5.01), (−3.29, −1.27), (0.8, 2.9), (5.09, 6.61).
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Optimization Problems
General steps:
1. Draw a picture and assign variables.
2. Write down the equation to be maximized or minimized (this is sometimes called the objective equation)
and the equation that describes the constraint (this is sometimes called the constraint equation).
3. Use the constraint equation to rewrite the objective equation so that it has only one independent variable.
4. Find the domain of the [new] objective equation.
5. Find the maximum or minimum using calculus. Technically we’re finding an absolute max/min, but in
these problems it very often occurs at a local max/min in the domain (and in many problems there’s
only one possibility).
6. Verify your answer using the Second Derivative Test.
Exercises
1. A farmer wants to build a pen with two dividers in order to separate elephants, donkeys, and penguins.
If 600 ft of fence is available and one side of the pen is bounded by a river and needs no fence since all
the animals just happen to have an irrational fear of water, then what is the maximum area that can
be enclosed?
2. A carpenter wants to build a rectangular box with square sides in which to put round things. The
material for the bottom costs $20/ft2 , material for the sides costs $10/ft2 , and the material for the top
costs $50/ft2 t. If the volume of the box must be 5 ft3 , then find the dimensions that will minimize the
cost (and find the minimum cost).
3. A knight sees a damsel in distress 3 miles downstream on the opposite side of a straight raging river 0.5
miles wide. The knight can swim at 4 mi/hr and run at 7 mi/hr. At what point on the opposite side
should the knight swim in order to reach the distressed damsel as soon as possible.
4. After being rescued the distressed damsel decided to buy a peach orchard, and she wants to maximize
the number of peaches produced by her orchard. She has found that the per-tree yield is equal to 900
whenever she plants 45 or fewer trees per acre, and that when more than 45 trees are planted per acre,
the per-tree yield decreases by 25 peaches per tree for every extra tree planted. For example, if there
were 40 trees planted per acre, each tree would produce 900 peaches. If there were 50 trees planted per
acre, each tree would produce 900 − 25(50 − 45) = 775 peaches. Find the number of trees that should
be planted per acre to maximize the yield, and find the maximum yield per acre.
5. A 5 in × 8 in piece of paper has a square cut out of each corner (same size from each) and is then folded
to make an open-top box. Find the size of the square that will maximize the volume.
6. √
Find √
the area of the largest rectangle that can be inscribed inside an isosceles triangle with side lengths
2, 2, 2.
7. A box with a square base and open top must have a volume of 32000 cm3 . Find the dimensions of the
box that will minimize the amount of material needed.
R
Bottling Company has asked you to design a cylindrical can that will hold 355 ml
8. Suppose Cheerwine
and uses the least amount of aluminium. What should be the dimensions of the can?
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